Show that a the sum function of a power series is differentiable twice and that $f''(x) = \frac{1}{49-7x}$

Question

I am studying for my analysis exam and are to consider the power series $$ \sum_{n=2}^\infty \frac{1}{n(n-1)7^n}z^n $$ with the sum function for $x \in ]-7,7[$ given by $$ f(x) = \sum_{n=2}^\infty \frac{1}{n(n-1)7^n}x^n $$ Then I have to show that $f$ is differentiable twice and that $$ f''(x) = \frac{1}{49-7x} $$ for $x \in ]-7,7[$. As I have shown earlier that $f$ is indeed differentiable as I have found that the convergence radius $R = 7 >0$ we have that $$ f''(x) = \sum_{n=2}^\infty \frac{d^2}{dx^2} \frac{1}{n(n-1)7^n}x^n = \sum_{n=2}^\infty \frac{n(n-1)}{n(n-1)7^n}x^{n-2} = \sum_{n=2}^\infty \frac{1}{7^n}x^{n-2} $$ but here I get stuck. I have the answer sheet and it says that we now can rewrite to $$ \sum_{n=2}^\infty \frac{1}{7^n}x^{n-2} = \frac{1}{7^2} \sum_{n=0}^\infty \left( \frac{x}{7} \right)^n $$ but I simply cannot see why. I tried to withdraw the first terms for $n=0$ and $n=1$ but it still made not sense to me. Do you mind explaining?

Answer 1

It is over apply sum of a g.p to it that is 1/1-x/7

Answer 2

$$ \sum_{n=2}^{\infty} \frac{x^{n-2}}{7^n} = \frac{1}{7^2} \cdot \sum_{n=2}^{\infty} \frac{x^{n-2}}{7^{n-2}} = \frac{1}{7^2} \cdot \sum_{n=0}^{\infty} \frac{x^{n}}{7^{n}} $$